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 A334765 Numbers m such that the numbers of 1's in the binary expansion of m equals the negative sum of balanced ternary trits of m. 2
 0, 41, 68, 131, 132, 368, 384, 528, 1095, 1098, 1100, 1106, 1112, 1122, 1124, 1152, 1176, 1346, 1824, 2561, 3282, 3284, 3336, 3344, 3392, 3524, 4098, 4101, 4104, 4112, 4118, 4128, 4172, 4352, 4496, 4739, 4740, 5504, 6224, 9856, 9857, 9869, 9896, 9923, 9924 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS a(116) = 32770, a(117) = 32771 and a(118) = 32772 is the first time that three consecutive numbers appear in this sequence.  Conjecture: There is no occurrence of four or more consecutive numbers.  Tested by exhaustive search up to a(n) = 3^26. - Thomas König, Jul 19 2020 LINKS Thomas König, Table of n, a(n) for n = 1..30000 FORMULA Integers m such that -A065363(m) = A000120(m). EXAMPLE 41_10 = 1TTTT_bt = 101001_2, the sum of the digits is 1-1-1-1-1 = -3 for balanced ternary and 1+1+1 = 3 for base 2, so 41 is a term. MATHEMATICA Select[Range[0, 10^4], -Total@ If[First@ # == 0, Rest@ #, #] &[Prepend[IntegerDigits[#, 3], 0] //. {x___, y_, k_ /; k > 1, z___} :> {x, y + 1, -1, z}] == DigitCount[#, 2, 1] &] (* Michael De Vlieger, Jul 08 2020 *) PROG (PARI) bt(n)= if (n==0, return (0)); my(d=digits(n, 3), c=1); while(c, if(d[1]==2, d=concat(0, d)); c=0; for(i=2, #d, if(d[i]==2, d[i]=-1; d[i-1]+=1; c=1))); vecsum(d); \\ A065363 isok(m) = bt(m) + hammingweight(m) == 0; \\ Michel Marcus, Jun 07 2020 CROSSREFS Aside from the first term, subsequence of A174657. Cf. A000120, A037301, A065363, A330904. Sequence in context: A039524 A259586 A140374 * A269807 A289982 A054806 Adjacent sequences:  A334762 A334763 A334764 * A334766 A334767 A334768 KEYWORD base,nonn AUTHOR Thomas König, May 10 2020 STATUS approved

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Last modified August 18 14:22 EDT 2022. Contains 356215 sequences. (Running on oeis4.)